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A-LevelMathematicsSeries9709/12 · May/June 2011 · Q1011 Marks

A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector.

  1. (a)A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector.[6]
  2. (b)(i)the value of the constant k,[3]
  3. (b)(ii)the sum to infinity of the progression.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

  1. [6] (a) a+5d = 4a or (a+4a)/2 × 6; B1 co; Sum of angles = 360° or 2π rads; (2a+5d)/2 × 6 = 360 or (a+4a)/2 × 6 = 360; M1 A1 Correct left-hand side. All correct.; Sim Eqns → a = 24° or 2π/15 rads; A1 Either answer.; Arc length = rθ = 5 * (2π/15); M1 Correct use of arc length with θ in rads.; Perimeter = rθ + 2r = 5 * (2π/15) + 2*5 = 12.094...; A1 co; Perimeter = 12.1.
  2. [3] (b)(i) k+6 / (2k+3) = k / (k+6); M1 A1 Correct eqn for k.; → k²-9k-36= 0 → k = 12; A1 Co condone inclusion of k = -3. (NB stating a, ar, ar² as f(k) gets M1)
  3. [2] (b)(ii) r = 18/27 = 2/3, a = 27; M1 A1 Correct formula for S∞ must have -1 < r < 1. co; → S∞ = 27/(1-2/3) = 81.

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About this question

Topic
Series, A-Level Mathematics (syllabus 9709). Structured question, 11 marks.
Source
A-Level Mathematics, May/June 2011 Paper 1 Variant 2, question 10.

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